Solution:
The goal is to express all terms with the same base (3) and use the Product Rule $a^m \times a^n = a^{m+n}$.
$$ 81^2 \times 3^{-7} = (3^4)^2 \times 3^{-7} $$
$$ = 3^{(4 \times 2)} \times 3^{-7} \quad \text{(Using Power of a Power Rule)} $$
$$ = 3^8 \times 3^{-7} $$
$$ = 3^{(8 + (-7))} \quad \text{(Using Product Rule)} $$
$$ = 3^1 $$
$$ = 3 $$
The value is 3.
Solution:
Express all bases as powers of prime numbers ($25=5^2$, $243=3^5$, $16=2^4$, $8=2^3$).
$$ \text{Expression} = \frac{(25)^{\frac{3}{2}} \times (243)^{\frac{3}{5}}}{(16)^{\frac{5}{4}} \times (8)^{\frac{4}{3}}} $$
$$ = \frac{(5^2)^{\frac{3}{2}} \times (3^5)^{\frac{3}{5}}}{(2^4)^{\frac{5}{4}} \times (2^3)^{\frac{4}{3}}} \quad \text{(Base transformation)} $$
$$ = \frac{5^{(2 \times \frac{3}{2})} \times 3^{(5 \times \frac{3}{5})}}{2^{(4 \times \frac{5}{4})} \times 2^{(3 \times \frac{4}{3})}} \quad \text{(Power of a Power Rule)} $$
$$ = \frac{5^3 \times 3^3}{2^5 \times 2^4} $$
$$ = \frac{5^3 \times 3^3}{2^{(5+4)}} \quad \text{(Product Rule in the denominator)} $$
$$ = \frac{125 \times 27}{2^9} $$
$$ = \frac{3375}{512} $$
The simplified value is $\frac{3375}{512}$.
Solution (Left Hand Side – L.H.S.):
Break down each term and use the Negative Exponent of a Fraction Rule $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$ and the Fractional Exponent Rule $\sqrt[n]{a^m} = a^{\frac{m}{n}}$.
Term 1: $\left(\frac{64}{125}\right)^{-\frac{2}{3}}$
$64 = 4^3$ and $125 = 5^3$.
$\left(\frac{64}{125}\right)^{-\frac{2}{3}} = \left(\left(\frac{4}{5}\right)^3\right)^{-\frac{2}{3}} = \left(\frac{4}{5}\right)^{(3 \times -\frac{2}{3})} = \left(\frac{4}{5}\right)^{-2} = \left(\frac{5}{4}\right)^2$
Term 2: $\frac{1}{\left(\frac{256}{625}\right)^{\frac{1}{4}}}$
$256 = 4^4$ and $625 = 5^4$.
$\frac{1}{\left(\frac{256}{625}\right)^{\frac{1}{4}}} = \frac{1}{\left(\left(\frac{4}{5}\right)^4\right)^{\frac{1}{4}}} = \frac{1}{\left(\frac{4}{5}\right)^{(4 \times \frac{1}{4})}} = \frac{1}{\left(\frac{4}{5}\right)^1} = \frac{5}{4}$
Term 3: $\frac{\sqrt{25}}{\sqrt[3]{64}}$
$\sqrt{25} = 5$
$\sqrt[3]{64} = \sqrt[3]{4^3} = 4$
$\frac{\sqrt{25}}{\sqrt[3]{64}} = \frac{5}{4}$
Combining the terms:
$$ \text{L.H.S.} = \left(\frac{5}{4}\right)^2 + \frac{5}{4} + \frac{5}{4} $$
$$ = \frac{25}{16} + \frac{5}{4} + \frac{5}{4} $$
$$ = \frac{25}{16} + \frac{10}{4} $$
$$ = \frac{25}{16} + \frac{10 \times 4}{4 \times 4} \quad \text{(To find a common denominator, 16)} $$
$$ = \frac{25}{16} + \frac{40}{16} $$
$$ = \frac{25 + 40}{16} $$
$$ = \frac{65}{16} $$
Since L.H.S. = $\frac{65}{16}$ and R.H.S. = $\frac{65}{16}$, the statement is proven.